What is the integral of (x^5+x+1)/(x^3-8) ?

The integral of (x^5+x+1)/(x^3-8) is

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integral of (x^5+x+1)/(x^3-8)

\int \frac{x ^{5} + x + 1}{x ^{3} - 8} d x

\frac{x ^{3}}{3} + \frac{35}{12} ln ∣ x - 2 ∣ + \frac{1}{12} (\frac{61}{2} ln \frac{1}{3} (x + 1)^{2} + 1 + 3 arctan (\frac{1}{3} (x + 1))) + C

Solution steps

\int \frac{x ^{5} + x + 1}{x ^{3} - 8} d x

Take the partial fraction of

\frac{x ^{5} + x + 1}{x ^{3} - 8} : x^{2} + \frac{35}{12 ( x - 2 )} + \frac{61 x + 64}{12 ( x ^{2} + 2 x + 4 )}

= \int x^{2} + \frac{35}{12 ( x - 2 )} + \frac{61 x + 64}{12 ( x ^{2} + 2 x + 4 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int x^{2} d x + \int \frac{35}{12 ( x - 2 )} d x + \int \frac{61 x + 64}{12 ( x ^{2} + 2 x + 4 )} d x

\int x^{2} d x = \frac{x ^{3}}{3}

\int \frac{35}{12 ( x - 2 )} d x = \frac{35}{12} ln ∣ x - 2 ∣

\int \frac{61 x + 64}{12 ( x ^{2} + 2 x + 4 )} d x = \frac{1}{12} (\frac{61}{2} ln \frac{1}{3} (x + 1)^{2} + 1 + 3 arctan (\frac{1}{3} (x + 1)))

= \frac{x ^{3}}{3} + \frac{35}{12} ln ∣ x - 2 ∣ + \frac{1}{12} (\frac{61}{2} ln \frac{1}{3} (x + 1)^{2} + 1 + 3 arctan (\frac{1}{3} (x + 1)))

Add a constant to the solution

= \frac{x ^{3}}{3} + \frac{35}{12} ln ∣ x - 2 ∣ + \frac{1}{12} (\frac{61}{2} ln \frac{1}{3} (x + 1)^{2} + 1 + 3 arctan (\frac{1}{3} (x + 1))) + C

y = (x^{2} + 3) ((2 x + 5)^{6})

\frac{d}{d x} (5 + \frac{1}{x ^{2}})

\int_{0}^{3} x^{2} - 4 d x

\int 29 x^{28} d x

x \to \frac{1}{3} lim (5 x - 7)

What is the integral of (x^5+x+1)/(x^3-8) ?
The integral of (x^5+x+1)/(x^3-8) is